Question: Is ${65241}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {65241}= &&{6}\cdot10000+ \\&&{5}\cdot1000+ \\&&{2}\cdot100+ \\&&{4}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {65241}= &&{6}(9999+1)+ \\&&{5}(999+1)+ \\&&{2}(99+1)+ \\&&{4}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {65241}= &&\gray{6\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {6}+{5}+{2}+{4}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${65241}$ is divisible by $9$ if ${ 6}+{5}+{2}+{4}+{1}$ is divisible by $9$ Add the digits of ${65241}$ $ {6}+{5}+{2}+{4}+{1} = {18} $ If ${18}$ is divisible by $9$ , then ${65241}$ must also be divisible by $9$ ${18}$ is divisible by $9$, therefore ${65241}$ must also be divisible by $9$.